Year 11: How to solve application of differentiation word problems

In Year 11 Mathematics, you often use differentiation to solve real-life optimisation problems. These are word problems where you are asked to maximise or minimise something, such as:

Maximum area or volume
Minimum cost
Minimum distance or time

Many students find these problems difficult, not because of the calculus, but because of the translation from words into algebra. This article gives you a clear “how to solve it” strategy that you can follow for almost any application of differentiation word problem.

Overall strategy: four stages

You can think of these word problems in four stages:

Understand the problem
Plan the solution
Carry out the plan
Reflect and interpret

We will go through each stage, then work through a full example step by step.

Stage 1: understand the problem

Step 1: read carefully and highlight key information

Read the question slowly. Underline or highlight:

What is given? (numbers, relationships, conditions)
What are the constraints? (fixed perimeter, fixed volume, etc.)
What quantity needs to be maximised or minimised?

Step 2: draw a diagram if possible

For geometry-style questions (rectangles, boxes, shapes, roads, fields), draw a simple sketch and label the important lengths or variables. The diagram does not need to be perfectly accurate.

Step 3: choose your variables

Decide what quantities will be your variables. Common choices:

Length and width of a rectangle: $x$ , $y$
Radius and height of a cylinder: $r$ , $h$
Time, distance or position: $t$ , $x$

Try to keep the number of variables small. Usually, you will end up with one independent variable in your final function.

Stage 2: plan the solution

Step 4: find a formula for the quantity to optimise

Write an expression for the thing you want to maximise or minimise. For example:

Area of a rectangle: $A = x y$
Surface area of a closed cylinder: $S = 2 π r ² + 2 π r h$
Volume of a rectangular box: $V = l w h$

This quantity is usually called $A (x)$ , $V (x)$ , $C (x)$ (for cost), or similar, depending on the context.

Step 5: use constraints to reduce to one variable

The word problem will usually give you a relationship between the variables, such as:

Perimeter is fixed: $2 x + 2 y = P$
Volume is fixed: $π r ² h = V$
Total length of material is fixed, etc.

Use this relationship to express one variable in terms of the other, then substitute into your expression from Step 4 so that you end up with a function of one variable only: $Q (x)$ .

Stage 3: carry out the plan (use differentiation)

Step 6: differentiate the function

Once you have a function like $Q (x)$ for the quantity you want to optimise, differentiate with respect to $x$ :

$Q' (x) = d Q / d x$

Step 7: find critical points

Solve the equation $Q' (x) = 0$ to find the critical points. These are the values of $x$ where the function could have a maximum or a minimum.

Also consider the endpoints of the domain if the variable must lie in a certain range (for example, a length cannot be negative).

Step 8: test for maximum or minimum

To check whether a critical point is a maximum or minimum, you can:

Use the second derivative test:
- If $Q'' (x) < 0$ , it is a local maximum.
- If $Q'' (x) > 0$ , it is a local minimum.
Or compare values of $Q (x)$ at the critical point(s) and at the endpoints.

Stage 4: reflect and interpret the answer

Step 9: interpret in context

Translate your mathematical solution back into the language of the problem:

Write a sentence that answers the actual question.
Include correct units (metres, square metres, dollars, etc.).

Step 10: check if the answer makes sense

Ask yourself:

Is the answer within realistic bounds? (no negative lengths or areas)
Does it satisfy the original constraint?
Would changing the value slightly make the quantity smaller or larger, as expected?

Worked example: maximum area of a fenced rectangle

The problem

A farmer has 80 metres of fencing to build a rectangular paddock. She will use all of the fence. What dimensions of the rectangle will give the maximum area?

Step 1: understand and define variables

Let the length of the rectangle be $x$ metres and the width be $y$ metres.

We are told:

The fence goes around all four sides.
Total fence length is 80 m.
We want to maximise the area ( $A = x y$ ).

Step 2: write the constraint

The perimeter of the rectangle is:

$2 x + 2 y = 80$

We can simplify by dividing everything by 2:

$x + y = 40$

Step 3: express one variable in terms of the other

From $x + y = 40$ , we can write:

$y = 40 - x$

Step 4: write the area as a function of one variable

The area is $A = x y$ . Substitute $y = 40 - x$ :

$A (x) = x (40 - x)$

Expand:

$A (x) = 40 x - x ²$

Step 5: find the derivative

Differentiate with respect to $x$ :

$A' (x) = 40 - 2 x$

Step 6: find critical points by solving A'(x) = 0

Set the derivative to zero:

$40 - 2 x = 0$

$2 x = 40$

$x = 20$

So the critical point is at $x = 20$ . Then:

$y = 40 - x = 40 - 20 = 20$

Step 7: check that this gives a maximum

Use the second derivative:

$A'' (x) = d / d x (40 - 2 x) = - 2$

Since $A'' (x) = - 2 < 0$ , the function is concave down, so the critical point is a maximum.

Step 8: interpret the solution

The rectangle has maximum area when it is a square with side length 20 m by 20 m.

The maximum area is:

$A (20) = 40 (20) - (20) ² = 800 - 400 = 400 m ²$

So, using 80 m of fence, the largest possible rectangular paddock is a 20 m by 20 m square with area 400 m².

Another example outline: minimum surface area of a can

Here is a brief outline for another common type of question. You can practise filling in the full details yourself.

The problem

A can is in the shape of a cylinder with fixed volume of 500 cm³. Find the radius and height of the can that minimise the material used, i.e. the total surface area.

Define variables:
Let radius be $r$ cm, height be $h$ cm.
Write the constraint (volume fixed):
$V = π r ² h = 500$ ⇒ $h = 500 / (π r ²)$ .
Write the quantity to minimise (surface area):
$S = 2 π r ² + 2 π r h$ .
Substitute $h$ :
$S (r) = 2 π r ² + 2 π r (500 / (π r ²)) = 2 π r ² + 1000 / r$ .
Differentiate:
$S' (r) = 4 π r - 1000 / r ²$ .
Find critical points:
Solve $4 π r - 1000 / r ² = 0$ for $r$ , then find $h$ using $h = 500 / (π r ²)$ .
Check minimum:
Use the second derivative or compare values to show it is a minimum.
Interpret the result:
Give $r$ and $h$ with appropriate units, and explain that this design uses the least material.

Common mistakes

Forgetting the constraint: Always write down the relationship (like perimeter or volume) and use it to reduce to one variable.
Not answering the question in words: At the end, write a sentence with units, such as “The maximum area is…” or “The dimensions that minimise the cost are…”.
Ignoring the domain: Check that your values are positive and make sense physically.
Stopping after finding the value of x: Often you must also find the corresponding value of $y$ or another quantity.

Summary checklist

When you see a differentiation word problem, ask yourself:

Have I defined my variables clearly?
Have I written the constraint equation?
Have I expressed the quantity to optimise as a function of one variable?
Have I differentiated and set the derivative equal to zero?
Have I checked that it is a maximum or minimum?
Have I written the final answer in a sentence with units and checked that it makes sense?

If you follow this “how to solve it” structure every time, differentiation word problems become much more systematic and less scary. With practice, you will start to recognise common patterns and solve them with confidence.